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The problem asks to calculate the pH of a solution where **537 mg of pure barium hydroxide (Ba(OH)₂)** is dissolved in **110 mL of water**. Here's a step-by-step breakdown:

### Step 1: Convert mass to moles of Ba(OH)₂
The molar mass of Ba(OH)₂ is:
\[
\text{Molar mass of Ba(OH)}_2 = 137.33 \, \text{g/mol (Ba)} + 2 \times (16.00 \, \text{g/mol (O)} + 1.008 \, \text{g/mol (H)}) = 171.34 \, \text{g/mol}
\]

Given mass = 537 mg = 0.537 g:
\[
\text{Moles of Ba(OH)}_2 = \frac{0.537 \, \text{g}}{171.34 \, \text{g/mol}} = 3.133 \times 10^{-3} \, \text{mol}
\]

### Step 2: Determine the concentration of OH⁻ ions
Ba(OH)₂ dissociates fully in water:
\[
\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-
\]

For every mole of Ba(OH)₂, there are **2 moles of OH⁻ ions**. Therefore, the moles of OH⁻ are:
\[
\text{Moles of OH}^- = 2 \times 3.133 \times 10^{-3} = 6.266 \times 10^{-3} \, \text{mol}
\]

Now, calculate the concentration of OH⁻ in the solution (volume is 110 mL = 0.110 L):
\[
[\text{OH}^-] = \frac{6.266 \times 10^{-3} \, \text{mol}}{0.110 \, \text{L}} = 0.0570 \, \text{mol/L}
\]

### Step 3: Calculate pOH
The pOH is calculated from the concentration of OH⁻ ions:
\[
\text{pOH} = -\log[\text{OH}^-] = -\log(0.0570) = 1.244
\]

### Step 4: Calculate pH
Since pH + pOH = 14 at 25°C:
\[
\text{pH} = 14 - \text{pOH} = 14 - 1.244 = 12.756
\]

Thus, the pH of the solution is **12.76** (rounded to 4 significant digits, matching the precision of the mass and volume given in the problem).

---

Would you like more details or have any questions about the steps? Here are some related questions for further exploration:

1. How would the pH change if a different volume of water was used?
2. What effect does temperature have on the dissociation of Ba(OH)₂ and pH?
3. How is the molarity of Ba²⁺ calculated in this scenario?
4. What would the pH be if a stronger base like NaOH were used instead?
5. How do significant digits affect the precision of pH calculations?

**Tip:** Always ensure you maintain consistency with significant figures in calculations, especially when dealing with logarithms in pH or pOH computations.

A chemist dissolves 537 mg of pure barium hydroxide in enough water to make up 110 mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.)

Learn how to calculate the pH of a barium hydroxide solution where 537 mg of Ba(OH)₂ is dissolved in 110 mL of water at 25 °C. The problem involves stoichiometric calculations, dissociation of Ba(OH)₂, and the relationship between pH and pOH. Suitable for high school chemistry students.

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